Question 13.3: Use Table 13.1 to integrate the following functions: (a) x^4......

(a) From Table 13.1, we find  \int x^n \mathrm{~d} x=\frac{x^{n+1}}{n+1}+c, n \neq-1 \text {. }  To find  \int x^4 \mathrm{~d} x \text { let } n=4  we obtain

\int x^4 d x=\frac{x^5}{5}+c

(b) From Table 13.1, we find  \int \cos (a x) \mathrm{d} x=\frac{\sin (a x)}{a}+c.  In this case a = k and so

\int \cos k x d x=\frac{\sin k x}{k}+c

Note that a, b, n and c are constants. When integrating trigonometric functions, angles must be in radians.

(c) From Table 13.1, we find  \int \sin (a x+b) \mathrm{d} x=\frac{-\cos (a x+b)}{a}+c.  In this case a = 3 and b = 2, and so

\int \sin (3 x+2) d x=\frac{-\cos (3 x+2)}{3}+c

(d) From Table 13.1, we find that if k is a constant then  \int k \mathrm{~d} x=k x+c .  Hence,

\int 5.9 d x=5.9 x+c

(e) In this example, the independent variable is t but nevertheless from Table 13.1 we can deduce

\int \tan (a t+b) \mathrm{d} t=\frac{\ln |\sec (a t+b)|}{a}+c

Hence with a = 6 and b = −4, we obtain

\int \tan (6 t-4) \mathrm{d} t=\frac{\ln |\sec (6 t-4)|}{6}+c

(f) The independent variable is z but from Table 13.1 we can deduce  \int \mathrm{e}^{a z} \mathrm{~d} z=\frac{\mathrm{e}^{a z}}{a}+c.  Hence, taking a = −3 we obtain

\int \mathrm{e}^{-3 z} \mathrm{~d} z=\frac{\mathrm{e}^{-3 z}}{-3}+c=-\frac{\mathrm{e}^{-3 z}}{3}+c

(g) Since  \frac{1}{x^2}=x^{-2},  we find

\int \frac{1}{x^2} \mathrm{~d} x=\int x^{-2} \mathrm{~d} x=\frac{x^{-1}}{-1}+c=-\frac{1}{x}+c

(h) When integrating  \cos 100 n \pi t  with respect to t, note that  100 n \pi  is a constant. Hence, using part (b) we find

\int \cos 100 n \pi t \mathrm{~d} t=\frac{\sin 100 n \pi t}{100 n \pi}+c